I've got a....

[Killswitch]

My personal favourite head-wrecking mathsy puzzle;

On a game show there are three doors behind one of which is the Star Prize. You choose the door you want to open and tell the host. Then he opens one of the doors you did not choose, showing you that it did not have prize behind it. At this point you can either stick with your original choice or change your mind and choose the other, still-closed door. What should you do?

It's such a famous one that I reckon a few people on here will know the answer so if you're going to explain then put it in spoiler tags for the ones who haven't seen this before. It's an interesting one.

 

[Logi]

My personal favourite head-wrecking mathsy puzzle;



It's such a famous one that I reckon a few people on here will know the answer so if you're going to explain then put it in spoiler tags for the ones who haven't seen this before. It's an interesting one.

oh i remember reading about that in an article, i found it rather mind blowing how deceiving that scenario is

 

[Access Denied]

Can't be arsed to read all the replies but BODMAS or otherwise surely the answer is 10.

If you apply BODMAS it's (10 x 0) + 10 or 0 + 10

If you work it out in the order it's written then it's still 10 because 10 + 10 x 0 is effectively 10 + 0.

Edit: Of course the answer could always be 20. But then it's 4 am and I need to sleep so I may be wrong about that last part...

 

[old.Tohtori]

I'd go with number 2, assassins don't just kill random people for the sake of it

Clearly the right answer

 

[gohan]

My personal favourite head-wrecking mathsy puzzle;



It's such a famous one that I reckon a few people on here will know the answer so if you're going to explain then put it in spoiler tags for the ones who haven't seen this before. It's an interesting one.

surely it's 50/50 at that point, assuming that the host knows what door it's behind

i.e it's behind 1 you choose 2 he shows you 3

you choose 1 he can show you either 2 or 3

you choose 3 and he shows you 2

in all of those scenarios it's then pure 50/50 as to wether you had it right to begin with, tho I'd probably swap as when you first chose it was 33% chance, it's now up to 50% so statistically more likely that you were wrong before one was eliminated.... but still 50/50 at the point of swapping

 

[TdC]

no, the trick is to *always* change doors. iirc it's statistics and probability.

http://plus.maths.org/content/solution-puzzle-no-4-three-doors

 

[gohan]

That's exactly what I said. Noob!!!

I guess if you did a load of tests on it and swaped everytime then the contestent should win 66% of the time, but it's not such a huge percentage that if it was a real life scenario you'd be comfortable swapping. If it's down to two doors and you have to choose (or re-choose) one then it's back to 50/50

 

[pcg79]

That's exactly what I said. Noob!!!

I guess if you did a load of tests on it and swaped everytime then the contestent should win 66% of the time, but it's not such a huge percentage that if it was a real life scenario you'd be comfortable swapping. If it's down to two doors and you have to choose (or re-choose) one then it's back to 50/50

no, its still three doors but you just dont have the choice of choosing one any more - ie the choices between 3 and 2 are not independent. so its never 50/50 as its still the same game - its always 1/3 chance per door. so always switch

 

[ileks]

The key to that 3 door puzzle is that the host knows which door has the prize behind it.

I know some really clever maths people that couldn't for the life of them work that one out. They got really pissed off about it

 

[Zenith.UK]

I like the 2 door riddle from "Labyrinth"

View: http://www.youtube.com/watch?v=2dgmgub8mHw

 

[Killswitch]

The key to that 3 door puzzle is that the host knows which door has the prize behind it.

I know some really clever maths people that couldn't for the life of them work that one out. They got really pissed off about it

Tell me about it...I understand the maths (just about) and I know the answer, but if I think about the scenario it still seems so wrong somehow that I have to resort to getting a pen out to prove it to myself!

Another fun (but less head-wrecking one is) this;

Three guests check into a hotel room. The clerk says the bill is $30, so each guest pays $10. Later the clerk realizes the bill should only be $25. To rectify this, he gives the bellhop $5 to return to the guests. On the way to the room, the bellhop realizes that he cannot divide the money equally. As the guests didn't know the total of the revised bill, the bellhop decides to just give each guest $1 and keep $2 for himself. Now that each of the guests has been given $1 back, each has paid $9, bringing the total paid to $27. The bellhop has $2. If the guests originally handed over $30, what happened to the remaining $1?

And the final one for this morning is a fun little maths question;

How many people need to be in a room before there is a 50% probability that two of them have the same birthday?

If you don't know that one, I will pretty much guarantee you'll get it very, very wrong.

 

[old.Tohtori]

One person with a multiple personality disorder, clearly.

 

[gohan]

Tell me about it...I understand the maths (just about) and I know the answer, but if I think about the scenario it still seems so wrong somehow that I have to resort to getting a pen out to prove it to myself!

Another fun (but less head-wrecking one is) this;



And the final one for this morning is a fun little maths question;



If you don't know that one, I will pretty much guarantee you'll get it very, very wrong.

the first one is just stupid, hard to explain it, but basically the wrong figures are being added, the new bill is £25 and thye pay 9 each which is 27 and the robbing fuck has the other £2, there is no missing £1.

the second one I assume 183?

as 366 will be 100% that at least 2 people share a birthday, so half that is 50% but tbh not really sure on the maths for that one

 

[ileks]

Completely wrong on the second one Gohan. Tat questions really goes to show how unintuitive probability can be.

 

[Ormorof]

is the second one 2?

 

[gohan]

just looked up the answer, and I understand the maths but it just seems so unrealistic...... but I guess when you think about it (I'e how many face book friends i have how many share birthdays with each other etc) then it is about right, but seems so bizare

 

[gohan]

is the second one 2?

lol no xD 2 people would have a 1/365 chance of sharing a birthday, it's once it gets to 3 or more that it become exponential